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\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
a. Đặt \(x^2-2y=a\)
ta có : \(\left(x^2-2y\right)^2-4\left(x^2-2y\right)-12=a^2-4a-12=a^2-6a+2a-12=\left(a-6\right)\left(a+2\right)\)
\(=\left(x^2-2y-6\right)\left(x^2-2y+2\right)\)
b. Đặt \(x+6=a\Rightarrow\left(x+3\right)\left(x+6\right)\left(x+9\right)+45=\left(a-3\right)a\left(a+3\right)+45\)
\(=a^3-9a+45\) nghiệm xấu quá không nhóm được ban ơi :((
\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)
\(=\left(x^2+6x\right)^2+8\left(x^2+6x\right)-9\)
\(=\left(x^2+6x+9\right)\left(x^2+6x-1\right)\)
\(=\left(x+3\right)^2\cdot\left(x^2+6x-1\right)\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
Đặt x^2-3x-2=t =>(t+4)(t-4)+12=t-16+12=t-4=(t+2)(t-2)
=>(x^2-3x-2+2)(x^2-3x-2-2)=(x^2-3x)(x^2-3x-4)
hi
sao bn ko viết cách làm ra cho mik bít lun